# How do these conditional probability tables work (resources)?

I've taken some probability classes before, but it's been a while. I'm looking for (1) some resource -- website, paper, software, book -- etc on conditional probabilities. Something which accessible to a junior/senior level undergrad. Specifically, I'm interested in the relationships described by the following example:

Suppose that the nodes $$A$$ and $$B$$ are independent, and the node $$C$$ depends on both (I understand this may be an example of what's called a directed acyclic graph, if that's important).
The type of dependencies associated with the nodes I described above rely on truth tables. For example, $$A$$ and $$B$$'s probability distributions:

$$D(A) = \begin{array}{c|c} \text{True} & \text{False} \\ \hline 0.3 & 0.7 \end{array}, \quad D(B) = \begin{array}{c|c} \text{True} & \text{False} \\ \hline 0.01 & 0.99 \end{array}$$

And $$C$$'s (in percentages)

Then, without observable evidence, the probability that $$C$$ is FALSE* is $$\approx 0.9688$$.
How does one arrive at that value?
Furthermore, how would I generalize this behavior to a larger graph, with possibly more complicated stuff like $$A$$ and $$B$$ not being independent.

• I´m not sure about the topic here. But I´ve calculated that the probability C is true is $\approx 1-0.9688=0.0312$ – callculus Sep 28 '18 at 14:00
• Have you checked your solution? – callculus Sep 28 '18 at 14:20
• Yep, sorry about that, I meant $P(C)$ is false. So you're right. – Zduff Sep 28 '18 at 14:30
• I have no recommendation for liturature. I can only show how to calculate $P(\overline C)$. – callculus Sep 28 '18 at 14:33
• Yep, that's all I need, if you don't mind. – Zduff Sep 28 '18 at 14:39

To calculate the probability that $$C$$ is false we calculate all combinations of $$A$$ und $$B$$ which lead to $$\overline C$$.

$$\overline C$$ is the complementary event of $$C$$.

$$P(\overline C)=P(A)\cdot P(B|A)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot P(B|\overline A)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot P(\overline B| A)\cdot P(\overline C|A\cap \overline B)+P(\overline A)\cdot P(\overline B|\overline A)\cdot P(\overline C|\overline A\cap \overline B)$$

The events are: $$A:=$$Event A is true, $$B:=$$Event B is true,$$\overline A:=$$Event A is false, $$\overline B:=$$Event B is false,$$C:=$$Event C is true, $$\overline C:=$$Event C is false

$$A$$ and $$B$$ are independent. Therefore $$P(B|A)=P(B)$$. This is the same for all combinations with the complementary events

$$P(\overline C)=P(A)\cdot P(B)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot P(B)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot P(\overline B)\cdot P(\overline C|A\cap \overline B)+P(\overline A)\cdot P(\overline B)\cdot P(\overline C|\overline A\cap \overline B)$$

And the conditional probability $$P(\overline C|A\cap B)=100 \%=1$$ is given in the table. This is the last row and third column. Similar for the combinations of the events A, B. Therefore

$$P(\overline C)=0.7\cdot 0.99\cdot 0.999+0.99\cdot 0.3\cdot 0.9+0.01\cdot 0.7\cdot 0.88+0.3\cdot 0.01\cdot 1 \approx 0.9688$$

And $$P(C)=1-P(\overline C)$$. This result can be also obtained by using the equivalent formula which is used to calculate $$P(\overline C)$$.